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Math Help - Differentiation - First Principles

  1. #16
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jen1603 View Post
    cancel the h with the h on the denominator ?

    but what about the h's that are inside square roots on the denominator ?
    note that those h's don't make the denominator zero. just the first h does, which you can cancel with the h in the numerator
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  2. #17
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    ok , thanks

    so now im ending up with

    -2

    /

    square root of 4x^2 x square root of 2x + square root of 2x


    how do i get my final answer ? and is that right ?
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  3. #18
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jen1603 View Post
    ok , thanks

    so now im ending up with

    -2

    /

    square root of 4x^2 x square root of 2x + square root of 2x


    how do i get my final answer ? and is that right ?
    assuming you meant

    \frac {-2}{\sqrt{4x^2} (\sqrt{2x} + \sqrt{2x})} (which is not actually what you typed)

    you are correct.

    note that (since x is positive), \sqrt{4x^2} = 2x, also, \sqrt{2x} + \sqrt{2x} = 2 \sqrt{2x}, and so you have

    \frac {-2}{2x \cdot 2\sqrt{2x}}

    now the 2's cancel and we have

    \frac {-1}{2x \cdot \sqrt{2x}}

    but \sqrt{2x} = (2x)^{\frac 12}, and so, 2x \cdot \sqrt{2x} = 2x \cdot (2x)^{\frac 12} = (2x)^{\frac 32}, which gives


    \frac {-1}{(2x)^{3/2}}

    or if you prefer

    \frac {-1}{(\sqrt{2x})^3}

    a quick check by differentiating \frac 1{\sqrt{2x}} = (2x)^{-1/2} using the chain rule confirms that this is the right answer.


    the simplifying may seem daunting, but it depends on how you do it. for example, it probably would have been best if we had not "simplified" the denominator in one of the steps to get \sqrt{4x(x + h)}. otherwise, with practice, such simplifications should be easy to do quickly
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